## Precalculus: Mathematics for Calculus, 7th Edition

The real solutions of the equation $3x^{2}+5x=2$ are $x=-2$ and $x=\frac{1}{3}$
1. Subtract 2 from each side: $3x^{2}+5x-2=0$ 2. Factor $3x^{2}+5x-2$. The factors are $(-x-2)(-3x+1)$ because $-x\times-3x=3x^{2}$(the a value in $a^{2}+bx+c$) and $-2\times1=-2$ the c value in the equation. When the factors are FOILED out, they equal the expression $3x^{2}+5x-2$. 3. Set the factors equal to zero: $-x-2 = 0$ and $-3x+1=0$ 4. Solve: $x=-2$ and $x=\frac{1}{3}$